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Chapter 7: Problem 36

Use the product-to-sum identities and the sum-to-product identities to find identities for each of the following. $$\cos \theta-\cos 7 \theta$$

Short Answer

Expert verified
\(\cos \theta - \cos 7\theta = -2 \sin(4\theta) \sin(3\theta)\)

Step by step solution

01

Identify the appropriate identity

We need to find the appropriate sum-to-product identity. For the expression \(\cos \theta - \cos 7\theta\), we use the identity: \[\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)\]\.
02

Assign variables

Assign \( A = 7\theta \) and \( B = \theta \) in the identity \( \cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \).
03

Substitute into the identity

Substitute \( A \) and \( B \) into the identity. This gives: \[\cos \theta -\cos 7\theta = -2 \sin\left(\frac{7\theta + \theta}{2}\right) \sin\left(\frac{7\theta - \theta}{2}\right)\]\.
04

Simplify the expression

Simplify the expression: \[\cos \theta - \cos 7\theta = -2 \sin\left(\frac{8\theta}{2}\right) \sin\left(\frac{6\theta}{2}\right) = -2 \sin(4\theta) \sin(3\theta)\]\.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

sum-to-product identities
Sum-to-product identities are used to convert the sum or difference of trigonometric functions into a product of trigonometric functions. This transformation can be very useful in simplifying expressions and solving trigonometric equations.
In our exercise, we have the expression \(\cos \theta - \cos 7\theta\). We used the identity:
  • \[\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)\]
By identifying which identity to use, and carefully substituting the appropriate angles, we simplify the original problem.
Always remember: finding the right identity is crucial for effective simplification.
product-to-sum identities
Product-to-sum identities, as the name suggests, help in converting products of sine and cosine functions into sums or differences. These identities are the reverse of sum-to-product identities and can often make complicated multiplications easier to handle. Although our specific example with \(\cos \theta - \cos 7\theta\) used a sum-to-product identity, understanding product-to-sum is equally important.
For instance, if you encounter a product like \(\cos A \cos B\), you can use the identity:
  • \[\cos A \cos B = \frac{1}{2} [\cos(A - B) + \cos(A + B)]\]
These transformations greatly aid in solving many trigonometric problems by breaking them down into more manageable pieces.
trigonometric simplification
Trigonometric simplification involves using identities to reduce trigonometric expressions to their simplest form. This process makes it easier to work with the expressions, whether for solving equations or evaluating them at specific points.

In our problem, we converted the difference of cosines into a product of sines using the identity:
  • \[\cos \theta - \cos 7\theta = -2 \sin(4\theta) \sin(3\theta)\]
By recognizing and applying the correct identity, we not only simplified the expression but made it more manageable for further manipulation or evaluation.
cosine difference
The cosine difference, \(\cos A - \cos B\), is a specific type of trigonometric expression that often appears in various mathematical problems. Understanding how to manipulate this difference is fundamental for solving many trigonometric problems.

In our example:
  • We recognize that \(A = 7\theta\) and \(B = \theta\).
  • We then applied the identity \[\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)\]
  • After substituting, we ended up with \[\cos \theta - \cos 7\theta = -2 \sin(4\theta) \sin(3\theta)\]
This method converts the difference of cosines into a more approachable form for further operations.

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Most popular questions from this chapter

The acceleration due to gravity is often denoted by \(g\) in a formula such as \(S=\frac{1}{2} g t^{2},\) where \(S\) is the distance that an object falls in time \(t .\) The number \(g\) relates to motion near the earth's surface and is generally considered constant. In fact, however, \(g\) is not constant, but varies slightly with latitude. Latitude is used to measure north-south location on the earth between the equator and the poles. If \(\phi\) stands for latitude, in degrees, \(g\) is given with good approximation by the formula \(g=9.78049\left(1+0.005288 \sin ^{2} \phi-0.000006 \sin ^{2} 2 \phi\right)\), where \(g\) is measured in meters per second per second at sea level. a) Chicago has latitude \(42^{\circ} \mathrm{N}\). Find \(g .\) b) Philadelphia has latitude \(40^{\circ} \mathrm{N}\). Find \(g\). c) Express \(g\) in terms of \(\sin \phi\) only. That is, eliminate the double angle.

Given that \(f(x)=\cos x,\) show that $$\frac{f(x+h)-f(x)}{h}=\cos x\left(\frac{\cos h-1}{h}\right)-\sin x\left(\frac{\sin h}{h}\right)$$

Solve, finding all solutions in \([0,2 \pi)\). $$\cos (\pi-x)+\sin \left(x-\frac{\pi}{2}\right)=1$$

Given that \(f(x)=\sin x,\) show that $$\frac{f(x+h)-f(x)}{h}=\sin x\left(\frac{\cos h-1}{h}\right)+\cos x\left(\frac{\sin h}{h}\right)$$

Acceleration Due to Gravity. (See Exercise 59 in Exercise Set \(7.2 .\) ) The acceleration due to gravity is e often denoted by \(g\) in a formula such as \(S=\frac{1}{2} g t^{2}\) -where \(S\) is the distance that an object falls in \(t\) seconds. The number \(g\) is generally considered constant, but En fact it varies slightly with latitude. If \(\phi\) stands for Jatitude, in degrees, an excellent approximation of \(g\) is eiven by the formula \(g=9.78049\left(1+0.005288 \sin ^{2} \phi-0.000006 \sin ^{2} 2 \phi\right)\) where \(g\) is measured in meters per second per second at sea level. At what latitude north does \(g=9.8 ?\)
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